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torisob [31]
4 years ago
11

????????????????????

Mathematics
2 answers:
Lyrx [107]4 years ago
8 0

Answer:

880cm^2

Step-by-step explanation:

adoni [48]4 years ago
3 0

im not sure if it is right but i got 920 cm2. is that one of ur answers?

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PQ⊥PS , m∠QPR=7x−9, m∠RPS=4x+22<br> Find : m∠QPR
emmainna [20.7K]

Given:

It is given that,

PQ ⊥ PS and

∠QPR = 7x-9

∠RPS = 4x+22

To find the value of ∠QPR.

Formula

As per the given problem PR lies between PQ and PS,

So,

∠QPR+∠RPS = 90°

Now,

Putting the values of ∠RPS and ∠QPR we get,

7x-9+4x+22 = 90

or, 11x = 90-22+9

or, 11x = 77

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or, x = 7

Substituting the value of x = 7 in ∠QPR we get,

∠QPR = 7(7)-9

or, ∠QPR = 40^\circ

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The value of ∠QPR is 40°.

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3 years ago
Solve for z<br> THANK U SM<br> I WILL MARK BRAINILIEST
Anit [1.1K]

Answer:

z=7

Step-by-step explanation:

Combine like terms: 2z=14

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Step by step please and thank you.
cupoosta [38]

Answer:

  the probability is 2/9

Step-by-step explanation:

Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).

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Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.

The joint probability of these two events occurring sequentially is the product of their probabilities:

  p(dime then nickel) = (2/5)×(5/9) = 2/9

_____

<em>Alternate solution</em>

You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.

  p(dime/nickel) = 20/90 = 2/9

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3 years ago
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The volume of the cylinder is 1,286.1.
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